A weak monobasic acid ( $K_{a} = 2.0 \times 1 0^{- 5}$ ) is $2.5%$ ionised in a certain solution. This solution is diluted with water to four times its original volume. Using the exact expression $K_{a} = c α^{2} / (1 - α)$ for both states, calculate the percentage ionisation after dilution, to two decimal places.

Numerical answer type, answer in %, tolerance 0.02.

Answer: 4.94 %

Initial concentration:

c_{1} = K_{a} (1 - α_{1}) / α_{1}^{2} = 2 \times 1 0^{- 5} \times 0.975/ (6.25 \times 1 0^{- 4}) = 0.0312 M

. After dilution

c_{2} = c_{1} /4 = 0.00780 M

. Solve

c_{2} α^{2} + K_{a} α - K_{a} = 0

α = 0.04937

, i.e.

4.94%

. Simple

α \propto 1/ c

would give exactly 5.00%, which is not exact.